1010 Radix (25分)

1010 Radix (25分)

Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true? The answer is yes, if 6 is a decimal number and 110 is a binary number.
Now for any pair of positive integers $N_1$ and $N_2$ ??, your task is to find the radix of one number while that of the other is given.

Input Specification:

Each input file contains one test case. Each case occupies a line which contains 4 positive integers:

N1 N2 tag radix

Here N1 and N2 each has no more than 10 digits. A digit is less than its radix and is chosen from the set { 0-9, a-z } where 0-9 represent the decimal numbers 0-9, and a-z represent the decimal numbers 10-35. The last number radix is the radix of N1 if tag is 1, or of N2 if tag is 2.

Output Specification:

For each test case, print in one line the radix of the other number so that the equation N1 = N2 is true. If the equation is impossible, print Impossible. If the solution is not unique, output the smallest possible radix.

Sample Input 1:

6 110 1 10

Sample Output 1:

2

Sample Input 2:

1 ab 1 2

Sample Output 2:

Impossible

题解：

#include <iostream>
#include <cctype>
#include <algorithm>
#include <cmath>
using namespace std;
long long convert(string n, long long radix) {
    long long sum = 0;int index = 0, temp = 0;for (auto it = n.rbegin(); it != n.rend(); it++) {
    temp = isdigit(*it) ? *it - '0' : *it - 'a' + 10;sum += temp * pow(radix, index++);}return sum;
}
long long find_radix(string n, long long num) {
    char it = *max_element(n.begin(), n.end());long long low = (isdigit(it) ? it - '0': it - 'a' + 10) + 1;long long high = max(num, low);while (low <= high) {
    long long mid = (low + high) / 2;long long t = convert(n, mid);if (t < 0 || t > num) high = mid - 1;else if (t == num) return mid;else low = mid + 1;}return -1;
}
int main() {
    string n1, n2;long long tag = 0, radix = 0, result_radix;cin >> n1 >> n2 >> tag >> radix;result_radix = tag == 1 ? find_radix(n2, convert(n1, radix)) : find_radix(n1, convert(n2, radix));if (result_radix != -1) {
    printf("%lld", result_radix);} else {
    printf("Impossible");}   return 0;
}